Calculus of a Single Variable, Chapter 9, 9.9, Section 9.9, Problem 7

A power series centered at c=0 is follows the formula:
sum_(n=0)^oo a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+...
The given function f(x)= 1/(1-3x) resembles the power series:
(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!) x ^n
or
(1+x)^k = 1+kx +(k(k-1))/(2!)x^2+(k(k-1)(k-2))/(3!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...
To evaluate the given function f(x) =1/(1-3x) centered at c=0 , we may apply Law of exponents: 1/x^n = x^(-n) .
f(x)= (1-3x) ^(-1)
Apply the aforementioned formula for power series on (1-3x) ^(-1) or (1+(-3x))^(-1) , we may replace "x " with "-3x " and "k" with "-1 ". We let:
(1+(-3x))^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) (-3x) ^n
=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)(-3)^nx ^n
=1+(-1)(-3)^1x +(-1(-2))/(2!)(-3)^2x ^2+(-1(-2)(-3))/(3!)(-3)^3x ^3+(-1(-2)(-3)(-4))/(4!)(-3)^4x ^4+...
=1+3x +2/2*9*x ^2+(-6)/6(-27)x ^3+24/24*81*x ^4+...
=1+3x +9x ^2+27x ^3+81x ^4+...
= sum_(n=0)^oo (3x)^n
To determine the interval of convergence, we may apply geometric series test wherein the series sum_(n=0)^oo a*r^n is convergent if |r|lt1 or -1 ltrlt 1 . If |r|gt=1 then the geometric series diverges.
By comparing sum_(n=0)^oo (3x)^n or sum_(n=0)^oo 1*(3x)^n with sum_(n=0)^oo a*r^n , we determine: r =3x .
Apply the condition for convergence of geometric series: |r|lt1 .
|3x|lt1
-1 lt3xlt1
Divide each part by 3:
(-1)/3 lt(3x)/3lt1/3
-1/3ltxlt1/3
Final answer:
The power series sum_(n=0)^oo(3x)^n has an interval of convergence: -1/3 ltxlt1/3 .